STRUCTURE OF Ti-44, Sc-44 AND Ca-44
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws in favor of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for discovering the nuclear force and structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Structure of unstable Ti-44 with S = 0 Comparing the unstable Ti-44 of S=0 with the stable Ca-40 of S=0 one concludes that Ti-44 consists of the two additional protons p21(+1/2) and p22(-1/2) as well as of the two additional neutrons n21(+1/2) and n22(-1/2) which contribute to the spin S=0 of Ti-44 giving the same S=0 as that of Ca-40. ( See my STRUCTURE OF Ca-40 AND Ca-48 ). For example Ca-40 (from p1 to n20) has 20 protons and 20 neutrons with S=0. Similarly the 22 protons and 22 neutrons of Ti-44 give the same total spin S= 0. In the following diagram one observes the additional deuteron of S = +1 having the p21(+1/2) and n21(+1/2). It belongs to the nucleons of fifth horizontal plane having positive spins, while the deuteron of p22 and n22 belongs to the nucleons of the second horizontal plane having negative spins. Here the n21 fills the blank position formed by p10 and p16 in order to make the two bonds (n21-p10) and (n21-p16) able to overcome the nn repulsions of short range. However the p21 which fills the blank position formed by n9 and n16, makes the two bonds (p21-n9) and (p21-n16) unable to overcome the pp repulsions of long range. Similarly the two np bonds of n22 overcome the short-ranged nn repulsions, while the two pn bonds of p22 cannot overcome the pp repulsions of long range. Under this condition the p22 turns into a neutron for the formation of the structure of Sc-44 with 21 protons and 23 neutrons'. ' ' ' Structure of unstable Sc-44 with S = +2 Under the pp repulsions of long range the Ti-24 disintegrates by electron capture to excited levels in Sc-44 having the nuclear structure with the additional deuteron of p21and n21 and the new extra neutrons n22 and n23 which are not shown here. Comparing the structure of Sc -44 with the structure of Ti-44 one concludes that the Sc-44 has the nucleons from p1 to n21 including the additional neutrons n22 and n23. Since the Sc-44 has S =+2 one concludes that the p22(-1/2) of Ti-24 here turns to the new neutron n23(+1/2) which fills the blank position formed by p9 and p14. Also the n22(-1/2) of Ti-44 here is moved as n22(+1/2) to fill the blank position formed by the p10 and the 20p of the alpha particle existing behind the parallelepiped of Mg-24. Therefore the spin S =+2 of Sc-44 is due to the summation of p21(+1/2), n21(+1/2), n22(+1/2) and n23(+1/2). However the two pn bond of p21(+1/2) cannot overcome the pp repulsions of long range . Under this condition the p21 turns into a new neutron. So the structure of Sc-44 decays to the stable structure of Ca-44 with S =0 because the extra n21, n22, n23 and n24 filling four symmetrical blank positions between protons are able to make stable np bonds able to overcome the nn repulsions of short range. ' ' Structure of stable Ca-44 with S = 0 It is of interest to note that Ca-44 consists of the nucleons from p1 to n20 with S=0 including the four extra neutrons like n21(+1/2), n22(-1/2), n23(+1/2), and n24(-1/2) with a total spin S=0. In the analysis of the unstable Sc-44 we saw that under the unstable bonds (p21-n9) and (p21-n16) of Sc-44 the p21(+1/2) turns into the n22(-1/2) of Ca-44 filling the blank position for the formation of the two stable bonds like (n22-p4) and n22-p13). Moreover the symmetriacal position of n23(+1/2) is the same as that of Sc-44. Then in order to fill the fourth symmetrical blank position the n22(+1/2) of Sc-44 is moved as n24(-1/2) in order to fill the symmetrical blank position between p3 and p15. In other words the stable structure of Ca-44 is due to the transformation of the p21 and p22 of the unstable Ti-44 and Sc-44 into the neutrons filling four symmetrical positions for the stability of Ca-44. ' ' ' STRUCTURE OF UNSTABLE Ti-44 WITH S = 0' (The structure consists of six horizontal planes existing from +HP1 of positive spins to the -HP6 with negative spins.The nucleons p17, n17, p18, n18, p19, n19, p20 and n20 of the two α particles in front of Mg-24 and behind it are not shown here) ' ' ' p12..........n12' ' n11..........p11 -HP6' ' n10..........p10..........n21' ' p9............n9..........p21 +HP5' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 -HP4' ' p13..........n6............p6............n15' ' n13..........p5...........n5............p15 +HP3' ' n22.........p4............n4' ' p22..........n3............p3 -HP2' ' n2............p2' ' p1...........n1 +HP1' Category:Fundamental physics concepts